Question: Let $a,$ $b,$ $c$ be distinct, nonzero real numbers such that
\[a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}.\]Find $|abc|.$

Note: Intermediate Algebra writing problem, week 12.
Solution: From the given equations,
\begin{align*}
a - b &= \frac{1}{c} - \frac{1}{b}  =\frac{b - c}{bc}, \\
b - c &= \frac{1}{a} - \frac{1}{c} = \frac{c - a}{ac}, \\
c - a &= \frac{1}{b} - \frac{1}{a} = \frac{a - b}{ab}.
\end{align*}Multiplying these equations, we get
\[(a - b)(b - c)(c - a) = \frac{(a - b)(b - c)(c - a)}{a^2 b^2 c^2}.\]Since $a,$ $b,$ and $c$ are distinct, we can cancel the factors of $a - b,$ $b - c,$ $c - a,$ to get
\[a^2 b^2 c^2 = 1.\]Therefore, $|abc| = \boxed{1}.$